Problem: $\left(3y^2+2\right)\dfrac{dy}{dx}=1$ and $y(-1)=1$. What is $x$ when $y=2$ ? $x=~$
Explanation: The differential equation is separable. What does it look like after we separate the variables? $\left(3y^2+2\right) dy=dx$ Let's integrate both sides of the equation. $\int \left(3y^2+2\right) dy=\int dx$ What do we get? $y^3+2y = x+C$ What value of $C$ satisfies the initial condition $y(-1)=1$ ? Let's substitute $x=-1$ and $y=1$ into the equation and solve for $C$. $\begin{aligned} 1^3+2\cdot1 &= -1+C\\ \\ 3 &= -1+C\\ \\ C&=4 \end{aligned}$ Now use this value of $C$ to find $x$ when $y=2$. $\begin{aligned} 2^3+2\cdot2&=x+4\\ \\ 8+4&=x+4\\ \\ x&=8 \end{aligned}$